Past exam of the mathematics course of the University of Cambridge / 2025 / ib / Paper 1 / 18H (Statistics)

Table of contents
- a 18H
  - Solution a
- b 18H
  - Solution b
- c 18H
  - Solution c
- d 18H
  - Solution d

a

Solution

Let

S_{xx} = \sum X_{i}^{2}

. The likelihood equations give

\hat{β} = \frac{\sum X _{i} Y _{i}}{S _{xx}}, \overset{σ}{^}^{2} = \frac{1}{n} \sum (Y_{i} - X_{i} \hat{β})^{2} .

These are orthogonal projections of a Gaussian vector onto the span of

X

and its orthogonal complement, hence are independent.

b

Solution

With

s^{2} = \sum (Y_{i} - X_{i} \hat{β})^{2} / (n - 1)

,

\frac{β ^ - β}{s / S _{xx}} \sim t_{n - 1} .

Thus the interval is

\hat{β} \pm t_{n - 1, 1 - α /2} \frac{s}{S _{xx}} .

c

Solution

Unbiasedness requires

\sum c_{i} X_{i} = 1

. Cauchy–Schwarz gives

1 \leq (\sum c_{i}^{2}) S_{xx}

, so

Var (\tilde{β}) = σ^{2} \sum c_{i}^{2} \geq \frac{σ ^{2}}{S _{xx}} .

Equality holds exactly for

c_{i} = X_{i} / S_{xx}

.

d

Solution

The reverse-regression estimate is

\hat{b} = \sum X_{i} Y_{i} / \sum Y_{i}^{2}

. Therefore

\hat{b} \hat{β} = \frac{( \sum X _{i} Y _{i} ) ^{2}}{( \sum X _{i}^{2} ) ( \sum Y _{i}^{2} )} \leq 1

by Cauchy–Schwarz. Equality means the two data vectors are proportional, which is exactly when both fitted residual sums, and hence both variance MLEs, vanish.

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